1. ## Coordinate Geometry

In the diagram above, triangle $\displaystyle ACD$ is such that the coordinates of $\displaystyle A$ and $\displaystyle B$ are $\displaystyle (2, 12)$ and $\displaystyle (6, 9)$ respectively. $\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC$. The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$. The equation of $\displaystyle AD$ is $\displaystyle 3y=5x+26$.

Calculate the coordinates of C.

2. Let $\displaystyle M_1(x_1,y_1), \ M_2(x_2,y_2)$ be two points and $\displaystyle M(x_M,y_M)$ a point on the segment $\displaystyle M_1M_2$ such as $\displaystyle \displaystyle\frac{M_1M}{MM_2}=k$.

Then $\displaystyle x_M=\displaystyle\frac{x_1+kx_2}{1+k}, \ y_M=\displaystyle\frac{y_1+ky_2}{1+k}$

In this case $\displaystyle \displaystyle\frac{AB}{AC}=\frac{1}{2}$

Then $\displaystyle x_B=\displaystyle\frac{x_A+\frac{1}{2}x_C}{1+\frac {1}{2}}$

$\displaystyle y_B=\displaystyle\frac{y_A+\frac{1}{2}y_C}{1+\frac {1}{2}}$

You know the coordinates of A and B, so you can find the coordinates of C.

3. Hello, Punch!

I assume there are more parts to this problem.
There is way too much information . . .

In the diagram, $\displaystyle \Delta ACD$ has points: $\displaystyle A(2,12)$ and $\displaystyle B(6,9).$

$\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC$.
The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$.
The equation of $\displaystyle AD$ is: $\displaystyle 3y\:=\:5x+26$.

Calculate the coordinates of $\displaystyle C.$

Here is a simplified diagram:

Code:
      |
|       A
| (2,12)o - - +
|        \    :
|         \   :
|          \  :
|           \ :
|            \:
|       B(6,9)o - - - - +
|              \        :
|               \       :
|                \      :
|                 \     :
|                  \    :
|                   \   :
|                    \  :
|                     \ :
|                      \:
|                     C o
|
- - + - - - - - - - - - - - - - -
|

$\displaystyle \text{Since }BC = \frac{2}{3}AC,\; BC\text{ is twice }AB$

Going from $\displaystyle A$ to $\displaystyle B$, we move: 4 units right and 3 units down.

So, going from $\displaystyle B$ to $\displaystyle C$, we move: 8 units right and 6 units down.

Therefore, point $\displaystyle C$ is $\displaystyle (14,3).$

4. you are right soroban